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User blog:Ikosarakt1/Introduction to pentational arrays
In this blog post I'll introduce my vision to how pentational arrays work. I shall start with structures only a bit larger that tetrational, then, when the main gist will be explained, I'll go to pentationals and probably even beyond. First, remember my blog post: formal definition of tetrational arrays. There I proposed to enclose structures into separators: \(A \&\ bp = \{b,p (A) 2\}\), where A is the any structure. It allows to make it more natural rather than use arrays in arrays. So, what is the \((X \uparrow\uparrow X)\) separator? I decided to use Bowers' ideas and used curly brackets. Let \((X \uparrow\uparrow X) = (\{X\}) = (\{X^X\}) = (\{X^{X^X}\}) = (\{X^{X^{X^X}}\}) = (\{\underbrace{X^{X^{\cdots^{X^X}}}}_{n\text{ }X's}\})\) In otherwords, there is no matter how high the power tower enclosed in curly brackets, because X can represent the arbitrary number of them. To distinguish it, however, from the other similar separators in the form \((X \uparrow\uparrow n)\), I've used curly brackets. So, \((\{X^X\}) = (X \uparrow\uparrow X)\), but \((X^X) = (X \uparrow\uparrow 2)\). When the array in the form \(\{b,p (\{X^{X^{\cdots^{X^X}}}\}) 2\}\), it is equal to \(X \uparrow\uparrow X \&\ bp\}\). This is the sketch of the new rule for my pentational arrays. All rules for tetrational arrays should be applied ignoring curly brackets. For example: \(\{3,3 (\{X^{X+1}\}) 2\} = \{3,3 (\{X^X\}*X\}) 2\} = \{3,3 (\{X^X\}*3\}) 2\} = \{3,3 (\{X^X\}+\{X^X\}+\{X^X\}) 2\}\) \(= \{\{X^X\} \&\ 3 (\{X^X\}) \{X^X\} \&\ 3 (\{X^X\}) \{X^X\} \&\ 3\} = \{X \uparrow\uparrow X \&\ 3 (\{X^X\}) X \uparrow\uparrow X \&\ 3 (\{X^X\}) X \uparrow\uparrow X \&\ 3\}\). Imagine that when it, eventually, after really long time, becomes \(\{3,N (\{X^X\}) \#\}\), we should construct \(X \uparrow\uparrow N \&\ 3\) (by my new rule), and N is very, very large. And we're barely scratch the surface of pentational arrays. The next natural question is: what is \(X \uparrow\uparrow (X+1)\) structure? Well, we can think about \(X+1\) as a row with X items (arbitrary number of them), and 1 item above (or below) it. Thus, we can use curly brackets as a row delimiters and \((X \uparrow\uparrow (X+1)) = ({\{X^{X^{\cdots^{X^X}}}\}}^X)\). That is, we can still apply the same rules (including my new, that I don't defined formally, but I believe that you understood how it work). For example: \(X \uparrow\uparrow (X+1) \&\ 32 = \{3,2 ({\{X^X\}}^X) 2\} = \{3,2 ({\{X^X\}}^2) 2\} = \{3,2 (\{X^{X*X}\}) 2\}\) \(= \{3,2 (\{X^{X*2}\}) 2\} = \{3,2 (\{X^{X+X}\}) 2\} = \{3,2 (\{X^{X+2}\}) 2\} = \{\{X^{X+1}\} \&\ 32 (\{X^{X+2}\}) \{X^{X+1}\} \&\ 32\}\) \(= \{3,3 (X) 3,3 (\{X^X\}) 3,3 (X) 3,3 (\{X^{X+1}\}) 3,3 (X) 3,3 (\{X^X\}) 3,3 (X) 3,3\}\). Using my rules, we can notate and solve the further structures: \(X \uparrow\uparrow (X+2) = }\}}^X}^X\) \(X \uparrow\uparrow (X+3) = }\}}^X}^X}^X\) \(X \uparrow\uparrow (X*2) = {\{X^{X^{\cdots^{X^X}}}\}}^{\{X^{X^{\cdots^{X^X}}}\}}\) \(X \uparrow\uparrow (X*2+1) = {\{X^{X^{\cdots^{X^X}}}\}}^ }\}}^X}\) \(X \uparrow\uparrow (X*3) = {\{X^{X^{\cdots^{X^X}}}\}}^ }\}}^{\{X^{X^{\cdots^{X^X}}}\}}}\) Now I really need to introduce some new type of separator that separates planes of X's: let normal curly brackets \(\{\}\) become \(\{\}^X\), and the new type should be \(\{\}^{X^2}\). Now \(X \uparrow\uparrow (X^2) = \{\cdots\cdots\}^{X^2}\), where \(\cdots\cdots\) just means \(X \times X\) grid of X's, X sections of X's separated by the normal curly brackets. For example: \(X \uparrow\uparrow X^2 \&\ 32 = \{3,2 ({\{X^X\}}^{X^X}) 2\} = \{3,2 ({\{X^X\}}^{X^2}\}) 2\} = \{3,2 ({\{X^X\}}^{X*X}\}) 2\} = \{3,2 ({\{X^X\}}^{X*2}\}) 2\}\), etc. Then we can make \(\{\}^{X^3}\)-typed brackets that separates realms, \(\{\}^{X^4}\)-typed brackets that separates flunes, and so on. Then you may ask, how to transform \(X \uparrow\uparrow A\) structure into a separator? Below is the process that shows how to convert \(X \uparrow\uparrow X^2 \&\ 32\) into \(\{3,2 ({\{X^X\}}^{X^X}) 2\}\): \(X \uparrow\uparrow X^2 \&\ 32 = X \uparrow\uparrow (X*X) \&\ 32 = X \uparrow\uparrow (X*2) \&\ 32\) \(= X \uparrow\uparrow (X+X) \&\ 32 = X \uparrow\uparrow (X+2) \&\ 32 = X^{X \uparrow\uparrow (X+1)} \&\ 32\) \(= {\{X^X\}}^{X \uparrow\uparrow X} \&\ 32 = {\{X^X\}}^{X \uparrow\uparrow 2} \&\ 32 = {\{X^X\}}^{X^X} \&\ 32\) \(= \{3,2 ({\{X^X\}}^{X^X}) 2\}\). In other words, we can simply evaluate all X's after \(\uparrow\uparrow\) to p's, solve the expression, make the power tower of that many X's, and then separate each \(p^A\) X's by \(\{\}^A\)-typed brackets. Notice that C can be also tetrational structure itself, for instance \(X \uparrow\uparrow (X \uparrow\uparrow (X+1))\), and we eventually can come to the point where one type of brackets expresses another. This is the limit of pentational arrays. We can also recall \(\{\}^A = \{\}^{X \uparrow\uparrow A}\) like I did similar replaces for tetrational arrays, and continue my hierarchy even further, for example, if \(X \uparrow\uparrow\uparrow X = \{X \uparrow\uparrow X \uparrow\uparrow X \cdots X \uparrow\uparrow X \uparrow\uparrow X\}^{X \uparrow\uparrow\uparrow X}\), then we can take \(X \uparrow\uparrow\uparrow (X+1) = \{X \uparrow\uparrow X \uparrow\uparrow X \cdots X \uparrow\uparrow X \uparrow\uparrow X\}^{X \uparrow\uparrow\uparrow X} \uparrow\uparrow X\). And we get larger and larger numbers. Category:Blog posts